Homework Answers
10.
(a)
Mole of propane = weight / MW = 4.63 g / 44.013 g/mole = 0.1052
mole
V = 500.0 mL = 0.500 L
T = 21 oC = 294 K
Ideal gas equation,
PV = nRT
P = nRT / V
= 0.1052 mole x 0.0821L.atm/mole.K x 294 K / 0.500 L
= 5.08 atm
(b)
Ideal gas equation,
PV = nRT
T = PV / nR
= 5.72 atm x 0.500 L / 0.1052 mole x 0.0821 L.atm/mole.K
= 331.14 K
= 58.14 oC
= 136.65 oF
(0°C × 9/5) + 32 = 32°F
(58.14 °C x 9/5) + 32 = 136.65 °F
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